Check whether 6^n cannot end with the digit 0 for any natural number n.

Fundamental Theorem of Arithmetic (FTA) states that every composite number can be uniquely expressed as a product of prime numbers.

Using FTA, we can say that any number ending with $0$ must have both $2$ and $5$ as prime factors. 

$6=2\times 3$

$(6{)}^n=(2\times 3{)}^n$

$(6{)}^n=(2{)}^n\times (3{)}^n$

Using FTA, we can say that $6^n$ does not have $5$ as a prime factor.

So, $6^n$ cannot end with the digit $0$ for any natural number $n$.